Em tham khảo nha:

\(\begin{array}{l}
{C_2}{H_4} + B{r_2} \to {C_2}{H_4}B{r_2}\\
C{H_4} + 2{O_2} \xrightarrow{t^0} C{O_2} + 2{H_2}O\\
2{C_2}{H_6} + 7{O_2} \xrightarrow{t^0} 4C{O_2} + 6{H_2}O\\
hh:{C_2}{H_4}(a\,mol),C{H_4}(b\,mol),{C_2}{H_6}(c\,mol)\\
\text{ Khối lượng tăng là khối lượng của $C_2H_4$ }\\
{n_{{C_2}{H_4}}} = \dfrac{{2,1}}{{28}} = 0,075\,mol\\
{n_{{H_2}O}} = \dfrac{{3,24}}{{18}} = 0,18\,mol\\
{n_{hh}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
\left\{ \begin{array}{l}
a + b + c = 0,15\\
a = 0,075\\
2a + 3b = 0,18
\end{array} \right.\\
 \Rightarrow a = 0,075;b = 0,045;c = 0,03\\
\% {V_{{C_2}{H_4}}} = \dfrac{{0,075}}{{0,15}} \times 100\%  = 50\% \\
\% {V_{C{H_4}}} = \dfrac{{0,045}}{{0,15}} \times 100\%  = 30\% \\
\% {V_{{C_2}{H_6}}} = 100 - 50 - 30 = 20\% 
\end{array}\)